Calculate the sun of the infinite series: \sum_{n=0}^\infty\frac{n}{4^n}

Sereinserenormg

Sereinserenormg

Answered question

2022-01-22

Calculate the sun of the infinite series:
n=0n4n

Answer & Explanation

hmotans

hmotans

Beginner2022-01-23Added 8 answers

Your sum is equal to
14+242+343+444++n4n+
Call this sum S. Now subtract from S the sum
14+142+143+
If we do it in the obvious way, term by term, we obtain
142+243+344+
Note that this last sum is (14)S
Putting things together, and using your computation for 1+14+142+ (not quite, we start at 1/4) we get
S13=S4
Solve for S. We find that S=49
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

A simple probabilistic approach: Let X be a geometric random variable with probability of success p, so that P(X=n)=(1p)n1p, n=1,2,3,... Then the expectation of X is E(X)=n=1nP(X=n)=n=1n(1p)n1p=p1pn=1n(1p)n Thus n=1n(1p)n=1ppE(X) On the other hand, from E(X)=p1+(1p)(1+E(X)) we get E(X)=1p Finally, n=1n(1p)n=1ppE(X)=1pp2 Letting p=3/4 gives n=1n4n=1/49/16=49
nick1337

nick1337

Expert2022-01-27Added 777 answers

First you can write n=1n4n=n=1m=114n Note that the above summation is over all (m,n) with mn. So if you switch the order of summation you obtain m=1n=m14n The inner sum is a geometric series with initial term 14m and ratio 14, so it sums to 14m114=4314m. So the overall sum is 43m=114m The sum here is a geometric series that sums to 13, so your final answer is 43×13=49

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